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最新推荐文章于 2022-03-07 19:58:40 发布

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本文介绍了一种算法，用于找出数组中所有两数之和为2的幂的组合。通过预计算2的幂并使用二分查找提高效率，确保了算法在大量数据下也能快速运行。

You are given n integers a₁, a₂, ..., a_n. Find the number of pairs of indexes i, j (i < j) that a_i + a_j is a power of 2 (i. e. some integer x exists so that a_i + a_j = 2^x).

Input

The first line contains the single positive integer n (1 ≤ n ≤ 10⁵) — the number of integers.

The second line contains n positive integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Output

Print the number of pairs of indexes i, j (i < j) that a_i + a_j is a power of 2.

Examples

Input

4
7 3 2 1

Output

Input

3
1 1 1

Output

Note

In the first example the following pairs of indexes include in answer: (1, 4) and (2, 4).

In the second example all pairs of indexes (i, j) (where i < j) include in answer.

题意找数组中有多少对数加起来是2的x次方思路给的数大小是1e9，和在2的30次方之内，二分法

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll x[35],a[101000];
int main()
{
    for(int i=1;i<=30;i++)
    {
        x[i] = pow(2,i);
        //cout<<x[i]<<endl;
    }
    ll n,ans,num;
    cin>>n;
    ans = 0;
    for(int i=0;i<n;i++)
    {
        scanf("%lld",&a[i]);
    }
    sort(a,a+n);
    for(int i=0;i<n;i++)
    {
        for(int j=1;j<=30;j++)
        {
            num = x[j] - a[i]; //二分法找符合的个数
            if(num<1||num>1e9)
                continue;
            int s = lower_bound(a+i+1,a+n,num)-a;//第一个等于num的位置
            int e = upper_bound(a+i+1,a+n,num)-a;//第一个大于num的位置
            ans += e-s;
        }
    }
    printf("%lld\n",ans);
    return 0;
}